Question

Is it possible using just one grep and regexp combination to achieve the following. Say I have a file like so:

$ cat f.txt
line 1 foo
line 2 boo
no match
line 3 blank
line X no match

I want to match all the lines that start with the word line and followed by a number but only display the what come after that, so the part that is matched by (.*).

$ grep -E '^line [0-9]+(.*)' f.txt 
line 1 foo
line 2 boo
line 3 blank

Can you say match but don’t display this part ^line [0-9]+ like doing the inverse of grep -o '^line [0-9]+'

So my expected output would look like this

$ grep -E ***__magic__*** f.txt 
foo
boo
blank

Answer 1

Given your example file:

$ cat cat_1.txt
line 1 foo
line 2 boo
no match
line 3 blank
line X no match

This is easy with Perl:

perl -lne 'print $1 if /^line \d+ (.*)/' cat_1.txt

Or with sed:

sed -En 's/^line [0-9]+ (.*)/\1/p' cat_1.txt

Either case, prints:

foo
boo
blank